Rewriting in terms of another (Trigonometry)

[LordofDirT]

Yay, finally our Trigonometry final is almost here. I'm taking the test this tuesday, and have been reviewing today.

One thing I've had a lot of trouble with is finding the correct sign on problems where I have to rewrite a trigonometric identity in terms of another.

For example:

Write tanx in terms of sinx, x is in Quadrant 4.

tanx = -sinx/cosx (-sinx because sinx is negative in Q4)

tanx = -sqrt(-sinx/sqrt(1-sin²x)

I am assuming sinx is negative because it is in Quadrant 4. Yet the answer in the book says the fraction is positive.

Yet...
Write tanx in terms of cosx; x in Quadrant 2:
this should have negative cosine values so

tanx = -sqrt(1-cos²x)/cosx is a correct answer in the back of the book.

Can anyone clarify anything about rewriting these with the correct sign values?

 

[Focus]

Ok from $$tan(x)= \frac{-sin(x)}{cos(x)}$$ try using a formula for $$sin^2(x)$$ (hint the one that involves 1 and $$cos^2(x)$$).

Good luck with your exam, I'm sure you will ace it.

 

[yeongil]

Write tanx in terms of sinx, x is in Quadrant 4.

tanx = -sinx/cosx (-sinx because sinx is negative in Q4)

tanx = -sqrt(-sinx/sqrt(1-sin²x)

I am assuming sinx is negative because it is in Quadrant 4. Yet the answer in the book says the fraction is positive.

Are you sure the part in bold above is correct? I'm getting a different answer.
The sin(x) is negative in Q4, that's right. But, in Q4,
* what is the sign of cos(x)?
* what is the sign of tan(x)?
* what is the sign of -sin(x)?
See if that helps.

Write tanx in terms of cosx; x in Quadrant 2:
this should have negative cosine values so

tanx = -sqrt(1-cos²x)/cosx is a correct answer in the back of the book.

Are you saying that you thought the answer is -sqrt(1-cos²(x))/-cos(x)? Check the signs of sin(x), tan(x), and -cos(x) in Q2.


01

 

[HallsofIvy]

Yay, finally our Trigonometry final is almost here. I'm taking the test this tuesday, and have been reviewing today.

One thing I've had a lot of trouble with is finding the correct sign on problems where I have to rewrite a trigonometric identity in terms of another.

For example:

Write tanx in terms of sinx, x is in Quadrant 4.

tanx = -sinx/cosx (-sinx because sinx is negative in Q4)

tanx = -sqrt(-sinx/sqrt(1-sin²x)
I am assuming sinx is negative because it is in Quadrant 4. Yet the answer in the book says the fraction is positive.

First, by definition, tan(x)= sin(x)/cos(x). You do NOT have a "-" here because both sin(x) and tan(x) are negative in the fourth quadrant. Since cos(x)= sqrt(1- sin²(x), tan(x)= sin(x)/sqrt(1- sin^2(x). Because both tan(x) and sin(x) are negative in the fourth quadrant, you want the positive square root here.
I don't know how you got that second square root!

Yet...
Write tanx in terms of cosx; x in Quadrant 2:
this should have negative cosine values so

tanx = -sqrt(1-cos²x)/cosx is a correct answer in the back of the book.

Can anyone clarify anything about rewriting these with the correct sign values?

In the second quadrant, sin(x) is positive while cos(x) is negative so tan(x) is negative. Since sin(x) and cos(x) are of opposite sign now you need the negative square root (or, since cos(x) is negative, cos(x)= -sqrt(1- sin²(x))).
tan(x)= -sinx(x)/sqrt(1- sin²(x))

Again, I have no idea how you got that second square root.